Exponential and Logarithmic Equations |
Contents: This page corresponds to § 4.4 (p. 348) of the text.
Suggested problems from text:
p. 355 #13,17,21,25,27,29,31,51,53,57,59,81,89
Exponential Equations
Logarithmic Equations
Some exponential equations can be solved by using the fact that exponential functions are one-to-one. In other words, an exponential function does not take two different values to the same number.
Example 1.
3x = 9
3x = 32
The function f(x) = 3x is one-to-one, so it does not take two different values to 9, so x must equal 2.
x = 2
The equation in example 1 was easy to solve because we could express 9 as a power of 3. However, it is often necessary to use a logarithm when solving an exponential equation.
Example 2.
ex = 20
We are going to use the fact that the natural logarithm is the inverse of the exponential function, so ln ex = x, by logarithmic identity 1. We must take the natural logarithm of both sides of the equation.
ln ex = ln 20
Now the left hand side simplifies to x, and the right hand side is a number. It is approximately 2.9957.
x = 2.9957
Use a calculator to check the answer we found to the equation in example 2.
Example 3.
5x = 16 We will solve this equation in two different ways.
First Approach: We use the fact that log5 5x = x (logarithmic identity 1 again).
5x = 16
log5 5x = log5 16
x = log5 16
x = ln 16 / ln 5, by the change-of-base formula.
x = 1.7227 (approximately)
Second Approach: We will use the natural logarithm and property 3.
5x = 16 Take the natural logarithm of both sides.
ln 5x = ln 16
x ln 5 = ln 16
x = ln 16 / ln 5
x = 1.7227 (approximately)
We could have used any logarithm with the second approach. The second approach is the one that you see most often.
Use a calculator to check the answer we found to the equation in example 3.
Equations like that in the next example occur frequently in applications.
Example 4.
200 e0.07t = 500
We first isolate the exponential part by dividing both sides of the equation by 200.
e0.07t = 2.5
Now we take the natural logarithm of both sides.
ln e0.07t = ln 2.5
The left hand side simplifies to 0.07t, by logarithmic identity 1.
0.07t = ln 2.5
t = ln (2.5) / 0.07
t = 13.1 (approximately)
Solve the following equations and check the answers.
(a) 3x = 10
(b) 150 e0.05 t = 350
When solving exponential equations we frequently used logarithmic identity 1 because it involves applying a logarithmic function to "undo" the effect of an exponential function. When dealing with logarithmic equations we will use logarithmic identity 2 where an exponential function is applied to "undo" the effect of a logarithmic function.
Example 5.
2 log x = 12
We want to isolate the log x, so we divide both sides by 2.
log x = 6
Since log is the logarithm base 10, we apply the exponential function base 10 to both sides of the equation.
10log x = 106
By logarithmic identity 2, the left hand side simplifies to x.
x = 106 = 1000000
Example 6.
7 + 3 ln x = 15 First isolate ln x.
3 ln x = 8
ln x = 8/3
Now apply the exponential function to both sides.
eln x = e8/3
x = e8/3
This is the exact answer. If you use a calculator to evaluate this expression, you will have an approximation to the answer.
x is approximately equal to 14.39.
Check the answers found in examples 5 and 6.
Example 7.
ln (x + 4) + ln (x - 2) = ln 7
First we use property 1 of logarithms to combine the terms on the left.
ln (x + 4)(x - 2) = ln 7
Now apply the exponential function to both sides.
eln (x + 4)(x - 2) = eln 7
The logarithmic identity 2 allows us to simplify both sides.
(x + 4)(x - 2) = 7
x2 + 2x - 8 = 7
x2 + 2x - 15 = 0
(x - 3)(x + 5) = 0
x = 3 or x = -5
x = 3 checks, for ln 7 + ln 1 = ln 7.
x = -5 does not check, for when we try to substitute -5 for x in the original equation we are taking the natural logarithm of negative numbers, which is not defined.
So, x = 3 is the only solution.